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Proof step 6 from proof ple-01 (previous step)

0 < j, j0 = i0 ∨ i0 < j0 ⊦ i0 < j + j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < j, i0 < j0 ⊦ j0 = i0, i0 < j + j0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ 0 < i → i1 < i1 + i; (s)